## On the concavity of t-norms and triangular functions.(Spanish. English summary)Zbl 0567.26010

In this paper the author proves the following Theorem: The unique 1/2- concave function $$T: [0,1]\times [0,1]\to [0,1]$$ increasing in each variable which satisfy $$T(1,x)=T(x,1)=x$$ for every $$x\in [0,1]$$ is $$M(x,y)=Min(x,y).$$ Let us consider the set $$\Delta^+=\{F| F: {\bar {\mathbb{R}}}\to I,\quad F(0)=0,$$ F is increasing and continuous to the left$$\}$$. For a fixed number $$a\in {\mathbb{R}}_+$$ the function $$\epsilon_ a$$ is defined by $$\epsilon_ a(x)=0$$ if $$x\leq a$$ and $$\epsilon_ a(x)=1$$ if $$x>a$$. Let be $$\tau: \Delta^+\times \Delta^+\to \Delta^+$$ a continuous function, increasing in each variable which satisfy $$\tau (F,\epsilon_ o)=\tau (\epsilon_ 0,F)=F$$ for every $$F\in \Delta^+$$ and $$\tau (\epsilon_ a,\epsilon_ a)=\epsilon_ a$$ for every $$a\in [0,1]$$. Using the above mentioned result the author shows that $$\tau$$ is concave if and only if $$\tau =\pi_ M$$ where $$\pi_ M(F,G)(x)=Min(F(x),G(x)).$$
Reviewer: D.Andrica

### MSC:

 26B25 Convexity of real functions of several variables, generalizations

### Keywords:

concave t-norm; concave triangular function
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